Is every continuous action of a compact topological group closed? I am reading Bredon's Introduction to compact transformation groups, and came across the following result and proof on page 34:

Even though he writes "Recall our standing assumption that $X$ is Hausdorff," I cannot see where any properties of Hausdorffness of $X$ are used in this proof. I suspect it would be used to ensure that convergent nets in $X$ converge uniquely, but I do not see why this fact would be needed for this proof.
Is it true that the continuous action of a compact topological group on a topological space $X$ is always closed, regardless of whether $X$ is Hausdorff? If not, then where in the above proof is Hausdorffness used?
 A: Here is an elementary proof:
Let $C\subseteq G\times X$ be closed. Note that $x_0\notin \Theta(C)$ if and only if $\{(g,g^{-1}x_0)\mid g\in G\}\cap C = \emptyset$ which is if and only if $G\times \{(e,x_0)\}\subseteq \Phi^{-1}((G\times X)\setminus C)$ where $\Phi: (g,(h,x))\mapsto (hg^{-1}, gx)$. Now, let $x_0\notin \Theta(C)$. By our observation, $G\times \{(e,x_0)\}\subseteq \Phi^{-1}((G\times X)\setminus C)$ and since $C$ is closed, $\Phi^{-1}((G\times X)\setminus C)$ is a neighborhood of $G\times \{(e,x_0)\}$. Hence, (by the tube lemma), we can find neighborhoods $U$ of $e$ and $V$ of $x_0$ such that $G\times (U\times V) \subseteq \Phi^{-1}((G\times X)\setminus C)$. It is easy to see that $\Theta(U\times V)$ is a neighborhood of $x_0$ which does not intersect $\Theta(C)$.
A: Yes, it is true, and the following argument will surely convince you so.
Recall that a topological space $K$ is compact iff it is universally closed:
for every topological space $X$, the coordinate projection map $X\times K \to X$ is closed. Now, use the obvious fact that $G\times X\to G\times X$, $(g,x)\mapsto (g,gx)$, is a homoemorphism and realize the action map $G\times X \to X$ as a composition of this homeo with the coordinate projection map.
